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[Leetcode] 8. String To Integer(atoi) TypescriptAlgorithm 2023. 1. 24. 15:29728x90
Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).
The algorithm for myAtoi(string s) is as follows:
- Read in and ignore any leading whitespace.
- Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
- Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
- Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
- If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.
- Return the integer as the final result.
Note:
- Only the space character ' ' is considered a whitespace character.
- Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.
Example 1:
Input: s = "42" Output: 42 Explanation: The underlined characters are what is read in, the caret is the current reader position. Step 1: "42" (no characters read because there is no leading whitespace) ^ Step 2: "42" (no characters read because there is neither a '-' nor '+') ^ Step 3: "42" ("42" is read in) ^ The parsed integer is 42. Since 42 is in the range [-231, 231 - 1], the final result is 42.
Example 2:
Input: s = " -42" Output: -42 Explanation: Step 1: " -42" (leading whitespace is read and ignored) ^ Step 2: " -42" ('-' is read, so the result should be negative) ^ Step 3: " -42" ("42" is read in) ^ The parsed integer is -42. Since -42 is in the range [-231, 231 - 1], the final result is -42.
Example 3:
Input: s = "4193 with words" Output: 4193 Explanation: Step 1: "4193 with words" (no characters read because there is no leading whitespace) ^ Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+') ^ Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit) ^ The parsed integer is 4193. Since 4193 is in the range [-231, 231 - 1], the final result is 4193.
Constraints:
- 0 <= s.length <= 200
- s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.
Solution
1. string이 number보다 앞에 있을 때 0으로 반환
2. sign뒤에 공백 또는 문자가 올 경우 0으로 반환
3. 조건 1, 2에 해당하지 않은 문자열은 number뒤에 오는 string은 제거function validate(string: string): boolean { const stringIndex = string.search(/[^0-9|^\+|^\-|^\s]/); const signIndex = string.search(/[\+|\-]/); const numberIndex = string.search(/[0-9]/); const isValidString = stringIndex < 0 || numberIndex < stringIndex; const isValidSignPosition = signIndex >= 0 && signIndex < numberIndex ? numberIndex - signIndex === 1 : true; return isValidString && isValidSignPosition; } function convertToNumber(raw: string, string: string): number { const startDigitIndex = raw.search(/[0-9]/); const isNegative = raw.substring(0, startDigitIndex).includes('-'); const stringWithRemovedSign = string.replace(/[^0-9]/g, ''); const isZero = Number(stringWithRemovedSign) === 0; const result = isNegative && stringWithRemovedSign !== '0' && !isZero ? -Number(stringWithRemovedSign) : Number(stringWithRemovedSign); if (result > Math.pow(2, 31) - 1) { return Math.pow(2, 31) - 1; } if (result < Math.pow(-2, 31)) { return Math.pow(-2, 31); } return result; } function myAtoi(s: string): number { if (!validate(s)) { return 0; } const startDigitIndex = s.search(/[0-9]/); const startStringIndex = s.substring(startDigitIndex).search(/[^0-9]/); const filteredString = startStringIndex >= 0 ? s.substring(startDigitIndex, startStringIndex + startDigitIndex) : s.substring(startDigitIndex); const result = convertToNumber(s, filteredString); return result; }
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